# How do you solve the following linear system: 8x+3y-6z=4, x-2y-z=2, 4x+y-2z=-4 ?

Dec 8, 2015

Combine $2$ pairs of equations to eliminate 1 of the variables from 2 equations; then solve the 2 equations for each of the remaining variables.
$\textcolor{w h i t e}{\text{XXX}} \left(x , y , z\right) = \left(- 4 , 0 , - 6\right)$

#### Explanation:

Given:
[1]$\textcolor{w h i t e}{\text{XXX}} 8 x + 3 y - 6 z = 4$
[2]$\textcolor{w h i t e}{\text{XXX}} x - 2 y - z = 2$
[3]$\textcolor{w h i t e}{\text{XXX}} 4 x + y - 2 z = - 4$

To eliminate $x$ from 2 equations:

Multiply [2] by $8$
[4]$\textcolor{w h i t e}{\text{XXX}} 8 x - 16 y - 8 z = 16$
Subtract [4] from [1]
[5]$\textcolor{w h i t e}{\text{XXX}} 19 y + 2 z = - 12$

Multiply [3] by $2$
[6]$\textcolor{w h i t e}{\text{XXX}} 8 x + 2 y - 4 z = - 8$
Subtract [6] from [1]
[7]$\textcolor{w h i t e}{\text{XXX}} y - 2 z = 12$

Now we have two equations in two unknowns:
[5]$\textcolor{w h i t e}{\text{XXX}} 19 y + 2 z = - 12$
[7]$\textcolor{w h i t e}{\text{XXX}} y - 2 z = 12$

[8]$\textcolor{w h i t e}{\text{XXX}} 20 y = 0$
[9]$\textcolor{w h i t e}{\text{XXX}} y = 0$
Substituting $0$ for $y$ in [7] (could have used [5])
[10]$\textcolor{w h i t e}{\text{XXX}} \left(0\right) - 2 z = 12$
[11]$\textcolor{w h i t e}{\text{XXX}} z = - 6$
Substituting $\left(- 6\right)$ for $z$ and $0$ for $y$ in [2] (could have used [1] or [3])
[12]$\textcolor{w h i t e}{\text{XXX}} x - 2 \cdot \left(0\right) - \left(- 6\right) = 2$
[13]$\textcolor{w h i t e}{\text{XXX}} x = - 4$